组合数学常用公式,经验式总结(持续更新)

1. 基本公式:
   \[ {n \choose k} = {n \choose n - k} \\ Pascal三角形:{n \choose k} = {n - 1 \choose k - 1} + {n - 1 \choose k}\\ 恒等式:\sum {n \choose i} = 2 ^ n\\ 二项式定理: (x + y)^n = \sum_{i = 0}^{n} {n \choose i} x^i y ^ {n - i} \]

2. 经验式:

\[ (r + 1) ^ k = \sum_{i = 0}^{k} {k \choose i} r ^ i\\ k {n \choose k} = n {n - 1 \choose k - 1}, 证明: (n - 1) - (k - 1) = n - k\\ 在二项式定理里,令x = 1, y = -1:\\ \sum_{i = 0}^{n} {n \choose i} (-1) ^ n = 0\\ 移项:\sum_{i = 0, i ~is~ odd}^{n} {n \choose i} = \sum_{i = 0, i ~is~ even}^{n} {n \choose i} = 2 ^{n - 1}\\ \]
\[ \sum_{i = 0}^{n} i{n \choose i} = n * 2 ^ {n - 1}\\ 证明:取二项式定理:(1 + n) ^ n = \sum_{i = 0}^{n} {n \choose i}\\ 两边求导: n(1 + n)^{n - 1} = \sum_{i = 1}^{n} i{n \choose i}n ^ {i - 1}, 取x = 1\\ 对于\sum_{i = 0} ^{ n} i ^k {n \choose i} 求k次导即可\\ \]
\[ \sum_{i = 0}^{n} {i \choose k} = {n + 1 \choose k + 1} \\ \sum_{i = 0}^{k} {n + i \choose i} = {n + k + 1 \choose k}\\ 分别对Pascal公式的第一项和最后意向反复迭代展开即可. \]

1. 范德蒙特卷积:

   \[ \sum_{k = 0}^{n} {m1 \choose k}{m2 \choose n - k} = {m1 + m2 \choose n}\]

   考虑组合意义证明, \({m1 + m2 \choose n}\)是\(m1 + m2\)向上向右的步子中向上的步数是\(n\)的方案数.

   那么我们把式子展开, 在前\(m1\)步中, 我们走\(k\)步,那么在后面的\(m2\)步中,我们一定要走\(n - k\)步

   所以有\[\sum_{k = 0}^{n} {n \choose k} ^ 2 = {2n \choose n} \]

转载于:https://www.cnblogs.com/qrsikno/p/10170523.html